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(A^n+B^n)^(1/n) limit stuff

  1. Jul 4, 2008 #1
    If [tex]0<A<B[/tex] then

    [tex]
    B < (A^n + B^n)^{1/n} < 2^{1/n} B,\quad\quad n=1,2,3,\ldots
    [/tex]

    and by using knowledge [tex]2^{1/n}\to 1[/tex] as [tex]n\to\infty[/tex], we get

    [tex]
    \lim_{n\to\infty} (A^n + B^n)^{1/n} = B.
    [/tex]

    Now, I'm interested to know what can we say if [tex]-B<A<0<B[/tex]. With odd n the inequality

    [tex]
    B\leq (A^n + B^n)^{1/n}
    [/tex]

    is wrong, so there does not seem to be any reason to believe that the limit would still be B. But is the limit still B?

    For example, set A=-1, B=2. The numbers

    n, (-1)^n + 2^n, ((-1)^n + 2^n)^(1/n)

    turn out as follows.

    1, 1, 1
    2, 5, 2.2361...
    3, 7, 1.9129...
    4, 17, 2.0305...
    5, 31, 1.9873...
    6, 65, 2.0052...
    7, 127, 1.9978...

    The 2^n appears to be dominating the limit quite well. Is the limit of this 2 really? What about more general conditions for A and B?
     
    Last edited: Jul 4, 2008
  2. jcsd
  3. Jul 4, 2008 #2
    I presume you meant |B| > A > 0 > B. (A^n + B^n)^1/n does not converge. This is best seen if you consider odd and even integers separately. If n only assumes even value, the expression converges to -B. If n assumes odd values, it converges to B. In a convergent sequence, every subsequence converges to the same value.
     
  4. Jul 4, 2008 #3
    Sorry. I fixed the mistake with edit option.
     
  5. Jul 4, 2008 #4
    I just realized it looks quite clear when you put it like this

    [tex]
    (A^n + B^n)^{1/n} = B\underbrace{\Big(\underbrace{\big(\frac{A}{B}\big)^n}_{\to 0} + 1\Big)^{1/n}}_{\to 1} \to B,
    [/tex]

    but I still encountered some problems with the proof...


    If we know that the sequence of functions

    [tex]
    f_n:[1-\delta, 1+\delta]\to\mathbb{R},\quad f_n(x)=(1+x)^{1/n}
    [/tex]

    with fixed [tex]0<\delta <1[/tex], converges towards

    [tex]
    f:[1-\delta, 1+\delta]\to\mathbb{R},\quad f(x)=1
    [/tex]

    uniformly, then the proof should be done, right? Looks clear by picture. I'm not sure how the uniform convergence is proven properly.
     
    Last edited: Jul 4, 2008
  6. Jul 4, 2008 #5
    Judging from what you were asking earlier, what you are implying is that B is negative (and so is A). If that's the case, factoring out B as you did is incorrect; (B^n)^1/n is not the same as B for an even n.

    Yes, for every fixed x, f_n(x) does converge to 1 uniformly as n goes to infinity. However, we are interested in g_n(x) = K*f_n(x) where K = B if n is even and K = -B if n is odd. For any x in the said interval, g_n(x) does not converge as n goes to infinity (there is no limit).

    Your argument works only in the case B is positive, in which case factoring out B is correct.

    If I'm misunderstanding you, please pardon me.
     
  7. Jul 4, 2008 #6
    You must have been left confused by my original typo. Now the original post assumes -B < 0 < B, so B is positive.

    With fixed x the word uniform doesn't have a meaning. What is relevant is that the convergence is uniform in some environment of x=1.
     
  8. Jul 4, 2008 #7
    If B is positive, what is the question exactly?

    By uniform convergence, I understood that you meant that for any x' in [1 - delta, 1 + delta], f_n(x') would form a uniformly decreasing sequence.
     
    Last edited: Jul 4, 2008
  9. Jul 4, 2008 #8
    The questions is that is the limit

    [tex]
    \lim_{n\to\infty}(A^n+B^n)^{1/n} = B
    [/tex]

    correct for all [tex]0<|A|<B[/tex]. I already knew how to prove the case [tex]0<A<B[/tex], but the case [tex]-B<A<0<B[/tex] seemed to be more difficult. However, I believe I already understood how this is proven too, provided that we know the one uniform convergence result mentioned in the post #4.

    I don't know what it means for a sequence of numbers to converge uniformly or not uniformly. The sequence of functions [tex]f_n:[a,b]\to\mathbb{R}[/tex] converges uniformly towards a function [tex]f:[a,b]\to\mathbb{R}[/tex], if for all [tex]\epsilon>0[/tex] there exists a number [tex]N\in\mathbb{N}[/tex] so that

    [tex]
    n>N\quad\implies\quad \underset{x\in[a,b]}{\sup} |f_n(x)-f(x)| < \epsilon.
    [/tex]
     
  10. Jul 4, 2008 #9
    ... environment of x=0. The domains should have been [tex][-\delta,\delta][/tex] too, instead of [tex][1-\delta, 1+\delta][/tex].
     
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